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# Planning and Control (Project Management) - Coursework Example

Summary
Exercise # 5.1 – Estimating with PERT Beta Distribution Question # 1 Data: The equipment of the company installs in the following time durations: The best case = 12 hours, The worst case = 18 hours, The most typical case = 14 hours Part A: Expected Time = (The best case + 4 * (The most typical case) + the worst case) / 6; Expected Time = (12 + 4 * (14) + 18) / 6; Expected Time = 14.33 hours Part B: Standard Deviation = (the worst case – the best case) / 6; Standard Deviation = (18 – 12) / 6; Standard Deviation = 1; Part C: The part A of the question presents that the expected time of installing the equipment in the company is 14.33 hours, however, there were three parameters estimated t…

## Extract of sample"Planning and Control (Project Management)"

Download file to see previous pages After completing the both parts provide a range of possible duration of the equipment installation activity that is the activity can be completed minimum of 13.33 hours and maximum in 15.33 hours. Exercise # 5.1 – Estimating with PERT Beta Distribution Question # 2 Data: The cost for design phase of the software development is as follows: The best case / lowest cost = \$35,000, The worst case / highest cost = \$60,000, The most typical case / likely cost = \$44,000 Part A: Expected Cost = (The best case + 4 * (The most typical case) + the worst case) / 6; Expected Cost = (\$35,000 + 4 * (\$44,000) + \$60,000) / 6; Expected Cost = \$45,166.67 Part B: Standard Deviation = (the worst case – the best case) / 6; Standard Deviation = (\$60,000 – \$35,000) / 6; Standard Deviation = \$4,166.67; Part C: The part A of the question presents that the expected cost of the developing software application in the design phase which is equivalent to the \$45,166.67. ...
After completing the both parts provide a range of possible cost of the design phase that is from \$41,000.00/- to \$49,333.33/-. Exercise # 5.1 – Estimating with PERT Beta Distribution Question # 3 Data: From the previous records, the following numbers of carpenters are required to complete the job: The best case / fewer number of carpenters = 4 carpenters, The worst case / largest number of carpenters = 9 carpenters, The most typical case / likely number of carpenters = 6 carpenters Part A: Expected number of carpenters = (The best case + 4 * (The most typical case) + the worst case) / 6; Expected number of carpenters = (4 + 4 * (6) + 9) / 6; Expected number of carpenters = 6.17 Part B: Standard Deviation = (the worst case – the best case) / 6; Standard Deviation = (9 – 4) / 6; Standard Deviation = 0.83; Part C: The part A of the question presents the expected number of carpenters required to complete the job and after applying the formula of PERT beta the calculated value is 6.17. In the part B of the question the standard deviation of the expected number of carpenters is calculated that is equivalent to 0.83. After subtracting and adding the value of standard deviation from / to the expected number of carpenters, the resulting value would provide us the estimated number of carpenters required to complete the job and the range is from 5.34 to 7 carpenters. Exercise # 5.1 – Estimating with PERT Beta Distribution Question # 4 – Dangers allied with PERT Technique The utilization of PERT technique involves several dangers and risks that include but are not limited to the following: The PERT technique becomes complicated when ...Download file to see next pagesRead More
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